Answer:
The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance
Null hypothesis is accepted at 0.05 level of significance
A company has developed a training procedure is significantly improve scores
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given sample mean (x⁻ ) = 517
sample standard deviation (s) =90
Mean of the Population (μ) =500
<em>Null Hypothesis :- H₀ : </em> (μ) =500
Alternative Hypothesis : H₁ : μ ≠ 500
<u><em>Step(ii)</em></u>:-
Test statistic
![t = \frac{x^{-} -mean}{\frac{s}{\sqrt{n} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%5E%7B-%7D%20-mean%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)
![t = \frac{517 -500}{\frac{90}{\sqrt{100} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B517%20-500%7D%7B%5Cfrac%7B90%7D%7B%5Csqrt%7B100%7D%20%7D%20%7D)
t = 1.717
<em>Level of significance</em>
∝ = 0.05
t₀.₀₅ = 1.6604
The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance
<u><em>Conclusion</em></u>:-
Null hypothesis is accepted at 0.05 level of significance
A company has developed a training procedure is significantly improve scores